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Thermodynamics - Calculate the initial temperature of the iron.

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A 100g of hot iron rivets is heated to some temperature and then quickly transferred to 120g of water at 40°C in a copper calorimeter of mass 50g and specific heat capacity of 400J/(kg.°C).It is found that the final temperature of the mixture is 100°C and the water has lost20% of its mass to steam. If the iron has a heat capacity of 460 J/(kg.°C), calculate the initial temperature of the iron. (The latent heat of vapour is 2.26 m/kg)


    2. Relevant equations

    E= M x C x ∆T and E = M x L Are these the equations I am supposed to use?

    3. The attempt at a solution

    I don’t want an answer to this. There is lot of data in this question which is confusing me. Can anyone please break the question down for me into simpler form? Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 28, 2012 #2

    Delphi51

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    Heat lost by iron = heat gained by warming the water + heat used to make steam
     
  4. Feb 28, 2012 #3
    Thank you Delphi51. :) I really appreciate your help. In the question they have given mass and heat capacity of copper calorimeter. I don't know what I need that information for and what I am supposed to work out first. :confused: Confused.com
    I really want to work thisout myself but I am not gettin anywhere with this.
     
  5. Feb 28, 2012 #4

    Delphi51

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    I missed that! Good catch.

    Heat lost by iron = heat gained by water + heat gained by calorimeter + heat used to make steam
     
  6. Feb 28, 2012 #5
    Thank you so much. I know how to work it out now. This is going to be the last question (hopefully)
    In the question it says iron quickly tranferred to water at 40°C in a calorimeter of mass 50g....
    I will be using E=MxCxΔT for heat gained by water and calorimeter and add them together. Do I use that 40°C in this equation for both of them?
    Like 120x4.2x(100-40) for water and 50x400x(100-40) for the container?
     
  7. Feb 28, 2012 #6

    Delphi51

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    Yes, the 100-40 looks good for both. 4.2 for the heat capacity does not look right. I seem to remember 4200. Better check it.
     
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